Tribonacci Convolution Triangle

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CSUSB ScholarWorks

CSUSB ScholarWorks

Electronic Theses, Projects, and Dissertations Office of Graduate Studies

6-2019

Tribonacci Convolution Triangle

Tribonacci Convolution Triangle

Rosa Davila

Follow this and additional works at: https://scholarworks.lib.csusb.edu/etd

Part of the Other Mathematics Commons

Recommended Citation Recommended Citation

Davila, Rosa, "Tribonacci Convolution Triangle" (2019). Electronic Theses, Projects, and Dissertations. 883.

https://scholarworks.lib.csusb.edu/etd/883

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A Thesis

Presented to the

Faculty of

California State University,

San Bernardino

In Partial Fulfillment

of the Requirements for the Degree

Master of Arts

in

Mathematics

by

Rosa Davila

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A Thesis

Presented to the

Faculty of

California State University,

San Bernardino

by

Rosa Davila

June 2019

Approved by:

Joseph Chavez, Committee Chair

Madeleine Jetter, Committee Member

Rolland Trapp, Committee Member

Shawnee McMurran, Chair, Department of Mathematics

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Abstract

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Acknowledgements

First and foremost, I want to thank my family for pushing me to be the best version of myself. If it weren’t for them, I don’t think I would have made it this far. They have created an environment that a person can grow in. They always promoted bettering yourself weather it be through school, work, or even healthy hobbies. My main focus was to be a person they can brag about. That is just my thanks to them for being such a supportive family.

The next next group of people I would like to thank are the friends I have made here at school. The fact that we were all struggling together made this a greater bond. Some I used for support for when I couldn’t handle the workload, and for others, I was a person they can follow. I had one of my classmates tell me “Congratulations on finishing your classes! Now all you have to do is graduate so that I can follow in your footsteps.” That made me feel amazing knowing that my peers are looking to me for support the way I looked at the graduate students before me.

I would like to personally thank Lacey for everything she has helped me with. When we were in our undergraduate courses together, we helped each other with home-work, studying, and also take home quizzes. I somewhat forced a friendship with her so that we can push each other both in and out of school. Once we got to the graduate program, we became closer friends. She was the main person that would push me to my breaking point to bring out the best graduate student ever. If it weren’t for her, I don’t think I would have had a great experience here in school. I’m so happy knowing that this school has brought me to meet such a great person!

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Table of Contents

Abstract iii

Acknowledgements iv

List of Tables viii

List of Figures x

1 Introduction 1

1.1 Different Arrays . . . 1

1.1.1 Pascal’s Triangle . . . 1

1.1.2 Fibonacci Convolution Triangle . . . 5

1.1.3 Tribonacci Convolution Triangle . . . 6

2 Sequences 7 2.1 Fibonacci Numbers . . . 7

2.2 Tribonacci Numbers . . . 8

2.3 Pell Numbers . . . 9

3 Generating Functions 10 3.1 Introduction to Generating Functions . . . 10

3.2 Convolutions . . . 11

3.3 Pascal’s Generating Function . . . 13

3.4 Pell Generating Function . . . 14

3.5 Fibonacci Generating Function . . . 15

3.6 Fibonacci Convolutions . . . 15

3.7 Tribonacci Generating Function . . . 16

3.8 Tribonacci Convolutions . . . 17

4 Convolution Triangles 18 4.1 Fibonacci Convolution Triangles . . . 18

4.1.1 Sum of Pascal’s Triangle Rows . . . 19

4.1.2 Pascal Meets Fibonacci . . . 21

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4.1.4 Fibonacci meets Tribonacci . . . 26

4.2 Tribonacci Convolution Triangles . . . 27

4.2.1 Tribonacci Meets Fibonacci . . . 28

5 Our Findings 31 5.1 Fibonacci Convolution Triangle . . . 32

5.1.1 Row Generators . . . 32

5.1.2 Sum of Rows for the mth Term . . . 33

5.1.3 The Sum of Distinct Term for the mth Term . . . . 37

5.2 Tribonacci Convolution Triangle . . . 39

5.2.1 Row Generators . . . 40

5.2.2 Sum of the Rows for the mth Term . . . 41

5.2.3 The Sum Distinct Terms for themth Term . . . 44

6 Conclusion 47 6.1 Fibonacci Convolution Triangle . . . 47

6.2 Tribonacci Convolution Triangle . . . 48

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List of Tables

1.1 Pascal’s Triangle Left Justified . . . 5

1.2 Fibonacci Convolution Triangle . . . 5

1.3 Tribonacci Convolution Triangle . . . 6

2.1 Rabbit Problem . . . 8

4.1 Fibonacci Convolution Triangle . . . 19

4.2 Pascal’s Triangle Left Justified . . . 19

4.3 Pascal’s Triangle Left Justified . . . 20

4.4 Pascal’s Triangle Left Justified . . . 21

4.5 Pascal’s Triangle Left Justified and Offset by Two Entries . . . 21

4.6 Pascal’s Triangel Offset by Two Entries . . . 22

4.7 First Column of Fibonacci Convolution Triangle . . . 22

4.8 Second Column of Fibonacci Convolution Triangle . . . 23

4.9 Third Column of Fibonacci Convolution Triangle . . . 24

4.10 Fourth Column of Fibonacci Convolution Triangle . . . 24

4.11 Fibonacci Convolution Triangle . . . 25

4.12 Sums of Fibonacci Convolution Triangle Result in Pell Numbers . . . 26

4.13 Sum of Rows in Fibonacci Convolution Triangle are Tribonacci Numbers . 27 4.14 Tribonacci Convolution Triangle . . . 28

4.15 First Column of Fibonacci Convolution Triangle . . . 29

4.16 Second Column of Fibonacci Convolution Triangle . . . 29

4.17 Third Column of Fibonacci Convolution Triangle . . . 29

4.18 Fibonacci Convolution Triangle . . . 30

5.1 Term Coefficients for Row Generating Functions . . . 32

5.2 Sum of Rows . . . 34

5.3 Sum of Terms Offset by Two Rows . . . 37

5.4 Sum of Terms . . . 38

5.5 Sum of Rows . . . 41

5.6 Sum of Terms Offset by Three Rows . . . 44

5.7 Sum of Distinct Terms . . . 45

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List of Figures

1.1 Binomial Expansion . . . 2

1.2 Pascal’s Triangle . . . 2

1.3 Pascal’s Triangle (constructing third row) . . . 2

1.4 Pascal’s Triangle . . . 3

1.5 Pascal’s Triangle (constructing fourth row) . . . 3

1.6 Pascal’s Triangle . . . 3

1.7 Pascal’s Triangle (constructing fifth row) . . . 3

1.8 Pascal’s Triangle . . . 4

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Chapter 1

Introduction

The fun I find in Pascal’s Triangle is that there are many ways to construct the array. It is a very popular triangle with many applications that go along with it[KOS18]. The same is not said when you mention either the Fibonacci Convolution Triangle or the Tribonacci Convolution Triangle. The goal of this chapter is to get you to recall the information you might remember from different arrays and transition into the Convolution Triangles.

1.1

Different Arrays

We start out with one of the most popular arrays known to date. There have been many observations of patterns in Pascals Triangle and we will later see how it leads to the construction of the Fibonacci Convolution Triangle[KOS14]. We lastly follow it to both the Fibonacci Convolution Triangle, and the Tribonacci Convolution Triangle.

1.1.1 Pascal’s Triangle

Pascal’s Triangle is one of the most recognizable arrays because it is made up of numbers that are applicable to many areas of mathematics. This array is most rec-ognizable for its binomial expansion[KOS11]. These numbers are seen in areas of com-binatorics, statistics, and in the expansion of binomials. These numbers come from the binomial theorem:

(x+y)n= n X

k=1

n k

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An array can be constructed for each of the entries. For example, we can create a row of the entries n. The array below shows each of the entries for the first 7 rows of the binomial theorem. Notice, for each row labeled n has n+ 1 entries since we start with zero[KOS11].

Figure 1.1: Binomial Expansion 0 0 1 0 1 1 2 0 2 1 2 2 3 0 3 1 3 2 3 3 4 0 4 1 4 2 4 3 4 4 5 0 5 1 5 2 5 3 5 4 5 5 6 0 6 1 6 2 6 3 6 4 6 5 6 6

Recall that nr = r!·(nn!r)!. Following this equation, the result of the triangle above is the triangle below.

Figure 1.2: Pascal’s Triangle 1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

There is also another way to think of the same triangle. Start with ones on the outside, then add the two entries above. Following this pattern results in generating Pascal’s Triangle.

Figure 1.3: Pascal’s Triangle (constructing third row) 1

1 1

1 1+1 1

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Figure 1.4: Pascal’s Triangle 1

1 1

1 2 1

The next row will be a result of ones on the sides, and each other entry will be made up of the sum of the two numbers above it.

Figure 1.5: Pascal’s Triangle (constructing fourth row) 1

1 1

1 2 1

1 1+2 1+2 1

In this case, the sums will be 1+2 and 2+1 with ones on the outside.

Figure 1.6: Pascal’s Triangle 1

1 1

1 2 1

1 3 3 1

The next row is slightly more difficult since there are more entries, but you can see the pattern starting to form at this point. Again, the next row will be a result of the ones on the sides, and each other entry will be made up of the sum of the two numbers above it.

Figure 1.7: Pascal’s Triangle (constructing fifth row) 1

1 1

1 2 1

1 3 3 1

1 1+3 3+3 3+1 1

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Figure 1.8: Pascal’s Triangle 1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

If we continue this pattern on for a few more rows, we see that it is the same triangle we previously saw. Below, we see the same triangle constructed through the sums of the two numbers directly above each entry.

Figure 1.9: Pascal’s Triangle 1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

Note that with this rule, the kth term in thenth row is the sum of the two entries above it. Those two terms will be on the row that is one less than the term that we speak of. Those terms will also be the k−1th and the kth term. With this observation, we have stumbled upon Pascal’s Identity[KOS11].

Theorem 1.1. Pascal’s Identity

Let n and k be positive integers where k≤n. Then kn= nk11+ n−k1

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Table 1.1: Pascal’s Triangle Left Justified n

0

n

1

n

2

n

3

n

4

n

5

n

6

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1.1.2 Fibonacci Convolution Triangle

There are a few things we need to know about the Fibonacci convolution triangle before we start proving certain results about it. We need to know how to generate the array which involves the convolution of generating functions[KOS14]. We must first define generating functions[KOS18], convolutions[KOS18], and also come up with the generating function for the Fibonacci numbers themselves[KOS14]. We will learn about each of those parts separately in later chapters. For now, we should see that the following table shows the first five columns of the left-justified and offset by 1 row version of the Fibonacci convolution triangle[KOS14].

Table 1.2: Fibonacci Convolution Triangle F(0) F(1) F(2) F(3) F(4)

1

1 1

2 2 1

3 5 3 1

5 10 9 4 1

8 20 22 14 5

13 38 51 40 20

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1.1.3 Tribonacci Convolution Triangle

The same goes with the Tribonacci convolution triangle. Since we are comparing this triangle to the Fibonacci convolution triangle[KOS14], we will have to learn about that triangle first. We need to understand what a generating function is in order to understand the convolution of generating functions, then find the generating function to the Tribonacci sequence of numbers. Lastly, use the convolutions of those generating functions. For now, we should see that the following table shows the first five columns of the left-justified and offset by one row of the Tribonacci convolution triangle[KOS14].

Table 1.3: Tribonacci Convolution Triangle T(0) T(1) T(2) T(3) T(4)

1 1 1 1 1

1 2 3 4 5

2 5 9 14 20

4 12 25 44 70

7 26 63 125 220

13 56 153 336 646

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Chapter 2

Sequences

We see in this chapter the importance of sequences. The main point of this project is comparing the Fibonacci Convolution Triangle with the Tribonacci Convolution Triangle. These two triangles are made up of two separate sequences of numbers. It would only make sense that they follow a certain pattern which we will begin to see in this chapter.

2.1

Fibonacci Numbers

Fibonacci numbers have been seen in many different occurrences. One of the most popular of the bunch is the rabbit problem discoverd by Leonardo Fibonacci, which goes as follows[KOS18].

Suppose there are two new born rabbits, one male and the other female. Find the number of rabbits produced in a year if:

1) Each pair takes one month to become mature;

2) Each pair produces a female and male every month, from the second month on; and

3) No rabbits die during the course of the year.

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and one new pair of baby rabbits, since the old babies were not mature enough to birth a new pair, which leaves us with a total of three pair of rabbits. The table below is a visual version of the situation we are talking about[KOS18].

Table 2.1: Rabbit Problem

Pairs of Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Adults 0 1 1 2 3 5 8 13 21 34 55 89

Babies 1 0 1 1 2 3 5 8 13 21 34 55

Total 1 1 2 3 5 8 13 21 34 55 89 144

The numbers that make up the bottom row are called the Fibonacci numbers. The sequence of numbers 1, 1, 2, 3, 5, 8, 13, ... is the Fibonacci Sequence[KOS18] that follow the recursive definition of thenth Fibonacci number,

Fn =Fn−1 +Fn−2 for n ≥3 andF1=F2=1.

2.2

Tribonacci Numbers

Sadly there isn’t a famous story to help us come up with the Tribonacci numbers. However, the Tribonacci numbers[KOS18] are defined by the recurrence relation

Tn = Tn−1 +Tn−2 Tn−3 for n≥ 4 andT1=T2 = 1T3 = 2

The first three terms of the sequence are predetermined. The sequence adds the three consecutive terms to obtain the following term. We can see what they are below. We start to see the pattern being formed after the fourth term.

T1 = 1 T2 = 1 T3 = 2

T4 =T3 +T2 +T1 = 1 + 1 + 2 = 4 T5 =T4 +T3 +T2 = 1 + 2 + 4 = 7 T6 =T5 +T4 +T3 = 2 + 4 + 7 = 13 T7 =T6 +T5 +T4 = 4 + 7 + 13 = 24

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2.3

Pell Numbers

Pell numbers have an interesting recurrence that are similar to a few differ-ent sequences[KOS11]. For instance, the Pell sequence is very similar to the Pell-Lucas sequence[KOS11]. The only difference is the first two terms. For this project, we will focus on the Pell numbers[KOS11]. The sequence follows the following relation

Pn = 2·Pn−1 +Pn−2 forP1 = 1,P2= 2.

In order to start this sequence, we are given the first two terms. The sequence doubles the previous term, then adds the term before that. Below is a more organized version of the relation.

P1 = 1 P2 = 2

P3 = 2·P2+P1 = 2·2 + 1 = 5 P4 = 2·P3+P2 = 2·5 + 2 = 12 P5 = 2·P4+P3 = 2·12 + 5 = 29 P6 = 2·P5+P4 = 2·29 + 12 = 70 P7 = 2·P6+P5 = 2·70 + 29 = 169 P8 = 2·P7+P6 = 2·169 + 70 = 408

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Chapter 3

Generating Functions

When it gets to the proving portion of this project, we find it easier to solve the problems using generating functions[KOS14]. We will start this chapter out with defining a few terms along with a few examples. This leads to the convolution of generating functions which we will need to define and also prove the convolution triangles of different sequences. Lastly, we will find the generating functions for sequences we need to solve for. Let us first start with an introduction of generating functions[KOS18].

3.1

Introduction to Generating Functions

Generating functions are used for both finite and infinite sequences[KOS18]. Since we are looking at sequences of numbers, we are going to concentrate more on the infinite forms of generating functions.

Definition 3.1. Let a0, a1, a2, ... be a sequence of real numbers. Then the function

g(x) = a0 + a1x2 +a2x3 + ...

is called the generating function for the sequence{an}[KOS18].

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g(x) = 1 +x+x2+x3+· · ·

=

X

n=0 xn

= 1 +

X

n=1 xn

= 1 +x·

X

n=1 xn−1

= 1 +x·

X

n=0 xn

= 1 +x·g(x).

At this point, we have both sides in terms of g(x). Below, we have the steps that will explicitly solve forg(x).

g(x)−x·g(x) = 1 g(x)·(x−1) = 1

g(x) = 11x

If we were to extend this out, we see that this generating function creates the same as previously stated. We can say that this is the generating function for the sequence of 1’s[KOS18]. In this project, we talk about sequences rather than the generating function. Note that the coefficients of the generating function,a0,a1,a2, ..., are in fact the sequences of {an}. This will make for a more simple way to prove the arrays.

3.2

Convolutions

Now understanding what a generating function is, and seeing how it can be applied, we can start applying the convolution to the generating function. After all, this project is on the Tribonacci Convolution Triangle. Since we are constructing a convolution triangle based on a sequence of numbers, we must have a solid understanding of the term convolution and what it brings to the array[KOS14].

Definition 3.2. Let a(x) and b(x) be the generating functions for the sequences{an}and

{bn}, then:

a(x) =a0+a1x+a2x2+a3x3+... b(x) =b0+b1x+b2x2+b3x3+...

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a(x)·b(x) =a0b0+ (a0b1+a1b0)x+ (a0b2+a1b1+a2b0)x2+...

If we letcn be the coefficients ofxn as such, then:

c0 =a0b0 c1 =a0b1+a1b0 c2 =a0b2+a1b1+a2b0

.. .

cn=a0bn+a1bn−1+a2bn−2+...+an−2b2+an−1b1+anb1

Then the product of the sequences {an} and {bn}, {cn} is called the convolution of {an}

and {bn}

Note, when taking the convolution of{an} and{bn}, the coefficients of{cn} are the sum of the coefficients of specified products of {an} and {bn}[KOS14]. If {ai} = {bi} = 1, then we are looking at the sum of 1’s as shown below, which means{cn}=n, the natural numbers. This will become more apparent in the following section.

c0 = a0b0 c1 = a0b1+a1b0 c2 = a0b2+a1b1+a2b0

.. .

cn = a0bn+a1bn−1+a2bn−2+...+an−2b2+an−1b1+anb1

If {ai} ={bi} = 1, then we are looking at the sum of 1’s as shown below, which means

{cn}=n, the natural numbers. This will become more apparent in the following section.

c0 = 1·1 = 1

c1 = 1·1 + 1·1 = 2

c2 = 1·1 + 1·1 + 1·1 = 3 ..

.

cn = 1·1 + 1·1 +...+ 1·1 + 1·1 = n

Convolutions have another fascinating characteristic. For generating functions who’s first few constants are 0, we start to lose starting points as we take convolutions[KOS14]. For example. if we had the following functions

a(x) =a1x+a2x2+a3x3+... b(x) =b1x+b2x2+b3x3+...

Then the convolution of the sequences{an}and {bn},{cn}, would be c(x) =a1b1x2+ (a1b2+a2b1)x3+ (a1b3+a2b2+a3b1)x4+...

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function at x2, this would offset our convolutions by two terms. If we were to start our generating functions atx3, this would offset our convolutions by three terms. Continuing this pattern, if we were to start our generating functions at xn, this would offset our convolutions by n terms. This will help us out when it comes to proving our patterns later in this project.

3.3

Pascal’s Generating Function

Recall Pascal’s Triangle as left justified where each column starts one row lower than the previous column[KOS11]. Note, the left-most column is made up of all 1’s, the second column is made up of the natural counting numbers, the third column is made up of triangular numbers, and so on and so forth. In this section, we will find the generating function that makes up each column of Pascal’s Triangle. Let us start with a general generating functions {an} and {bn}.

a(x) =a1x+a2x2+a3x3+a4x4+a5x5+...and b(x) =b1x+b2x2+b3x3+b4x4+b5x5+... If we letai=bi=1, for i=1, 2, 3, 4, 5, ... , we have

a(x) =x+x2+x3+x4+x5+...and b(x) =x+x2+x3+x4+x5+...

It is easy to see a(x) is a great representative of the first column of Pascal’s Triangle as left justified, and it’s respective generating function is

a(x) = x+x2+x3+x4+x5+... = 1xx

Now, by definition, the convolution of the first column with itself, let’s sayc(x), is a(x)·b(x) = c(x)

= x2(a1b1) +x3(a1b2+a2b1) +x4(a1b3+a2b2+a3b1) +... = x(1) +x2(1 + 1) +x3(1 + 1 + 1) +x4(1 + 1 + 1 + 1) +... = x+ 2x2+ 3x3+ 4x4+ 5x5+...

which makes up the second column of Pascal’s Triangle[KOS18], with generating function c(x) = (1x2x)2.

The convolution of the first column,a(x), with the second column,c(x), let’s sayd(x), is a(x)·c(x) = d(x)

= x2(a1c1) +x3(a1c2+a2c1) +x4(a1c3+a2c2+a3c1) +... = x(1) +x2(1 + 2) +x3(1 + 2 + 3) +x4(1 + 2 + 3 + 4) +... = x+ 3x2+ 6x3+ 10x4+ 15x5+...

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d(x) = (1x3x)3.

The convolution for the first column, a(x), with the third column,d(x), let’s saye(x), is a(x)·d(x) = e(x)

= x2(a1·d1) +x3(a1·d2+a2·d1) +... = x(1) +x2(1 + 3) +x3(1 + 3 + 6) +... = x+ 4x2+ 10x3+ 20x4+ 35x5+...

which makes up the third column of Pascal’s Triangle[KOS18] with generating function e(x) = (1x4x)4.

At this point it is easy to see that the generating functions that make up the entire array is

gn(x) = (1xx)nn+1

where the input of nmakes up thenth column.

3.4

Pell Generating Function

One of the other sequences of numbers that we need to discuss in this chapter are the Pell Numbers[KOS11]. Note the sequence of Pell Numbers follows the definition

Pn= 2·Pn−1+Pn−2 for n≥3 and P1 = 1, P2 = 2. Let us start with a general generating function[KOS18]. Let

g(x) =

X

n=1 anxn

= a1x+a2x2+

X

n=3 anxn

= a1x+a2x2+

X

n=3

(2·an−1+an−2)xn

= a1x+a2x2+

X

n=3

2·an−1xn+

X

n=3

an−2xn

= a1x+a2x2+ 2x·

X

n=3

an−1xn−1+x2·

X

n=3

an−2xn−2

= a1x+a2x2+ 2x·[g(x)−a1x] +x2·g(x). Letting a1= 1 and a2= 2, we have

g(x)−2x·g(x)−x2·g(x) = a1x+a2x2−2·a1x2 g(x)·[1−2x−x2] = x+ 2x2−2x2

g(x) = 12xxx2

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3.5

Fibonacci Generating Function

As previously stated, generating functions are used a lot in this project because we can easily see them when we start proving the different patterns. In this section, we will find the generating functions that results in the sequence of Fibonacci numbers. We start with recalling the definition of Fibonacci numbers[KOS18].

Definition 3.3. The sequence of numbers 1, 1, 2, 3, 5, 8, 13, make up the Fibonacci

sequence and follow the recursive definition of the nth Fibonacci number,

Fn =Fn−1 + Fn−2 for n≥3 and F1=F2=1

Now, let us start with a general generating function[KOS18]. Let g(x) =

X

n=1 anxn

= a1x+a2x2+

X

n=3 anxn

= a1x+a2x2+

X

n=3

(an−1+an−2)xn

= a1x+a2x2+

X

n=3

an−1xn+

X

n=3

an−2xn

= a1x+a2x2+x·

X

n=3

an−1xn−1+x2·

X

n=3

an−2xn−2

= a1x+a2x2+x·[g(x)−a1x] +x2·g(x) = a1x+a2x2+x·g(x)−a1x2+x2·g(x)

.

Letting a1=a2 = 1, we have

g(x) = x+x2+x·g(x)−x2+x2·g(x) g(x)−x·g(x)−x2·g(x) = x+x2−x2

g(x)·[1−x−x2] = x g(x) = 1xxx2

which is the generating function for the Fibonacci sequence[KOS18].

3.6

Fibonacci Convolutions

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a(x) = 1xxx2.

The convolution of this function with itself, let’s say b(x), would be a(x)·a(x) = 1xxx2 ·1xxx2

b(x) = (1xx2x2)2

The convolution of the Fibonacci sequence withb(x), let’s sayc(x), would be a(x)·b(x) = 1xxx2 · x

2 (1−x−x2)2 c(x) = (1xx3x2)3

The convolution of the Fibonacci sequence withc(x), let’s say d(x), would be a(x)·c(x) = 1xxx2 · x

3 (1−x−x2)3 d(x) = (1xx4x2)4

At this point we can see a pattern starting to form. We can now say that gn(x) = x

n

(1−x−x2)n

is the generating function for thenth convolution of the Fibonacci numbers[KOS14].

3.7

Tribonacci Generating Function

The sequence of Tribonacci numbers were discussed in a previous section. Since this project is finding the similarities in patterns for the different arrays, we need to find the generating function of the Tribonacci numbers[KOS18]. Following the same steps as we have been in this chapter, we can start with the definition of Tribonacci numbers[KOS18]:

Tn=Tn−1+Tn−2+Tn−3 forn≥4 andT1=T2 = 1, T3 = 2. We can start with a general generating function. Let

g(x) =

X

n=1 anxn

= a1x+a2x2+a3x3+

X

n=4 anxn

= a1x+a2x2+a3x3+

X

n=4

(an−1+an−2+an−3)xn

= a1x+a2x2+a3x3+

X

n=4

an−1xn+

X

n=4

an−2xn+

X

n=4

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= a1x+a2x2+a3x3+x

X

n=4

an−1xn−1

+x2

X

n=4

an−2xn−2+x3

X

n=4

an−3xn−3

= a1x+a2x2+a3x3+x[g(x)−a1x−a2x2] +x2[g(x)−a1x] +x3g(x) Letting a1=a2 = 1 anda3 = 2, we have

g(x) = x+x2+x3+x[g(x)−x−x2] +x2[g(x)−x] +x3g(x)

g(x)[1−x−x2−x3] = x

g(x) = 1xxx2x3

which is the generating function for the Tribonacci sequence[KOS18].

3.8

Tribonacci Convolutions

As previously stated, we know that the function that generates the Tribonacci numbers[KOS18] is the function

a(x) = 1xxx2x3.

The convolution of this function with itself, let’s say b(x), would be a(x)·a(x) = 1xxx2x3 ·1xxx2x3

b(x) = (1xxx22x3)2

The convolution of the Tribonacci sequence with b(x), let’s say c(x), would be a(x)·b(x) = 1xxx2x3 · x

2 (1−x−x2x3)2 c(x) = (1xxx32x3)3

The convolution of the Tribonacci sequence with c(x), let’s say d(x), would be a(x)·c(x) = 1xxx2x3 · x

3 (1−x−x2x3)3 d(x) = (1xxx42x3)4

At this point we can see a pattern starting to form. We can now say that gn(x) = (1xxxn2x3)n

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Chapter 4

Convolution Triangles

At this point, we can start putting all the pieces together to the puzzle that is the Fibonacci Convolution Triangle, and the Tribonacci Convolution Triangle[KOS14]. We talked about the sequences of both Fibonacci numbers and Tribonacci numbers[KOS18], which lead to the generating functions of the sequences[KOS18], lastly the convolution of each[KOS14]. We can now construct each of the arrays with their respective parts. Let us start with the Fibonacci Convolution Triangle, along with some of the observations others before me have discovered[HB72][KOS14].

4.1

Fibonacci Convolution Triangles

The first, left-most column is the entries of Fibonacci Numbers, let’s say Fn(0). The next column results in the convolution of Fn(0) and Fn(0), let’s say Fn(1). The third column results in the convolution of Fn(0) and Fn(1), denoted Fn(2). The fourth column results in the convolution ofFn(0) andFn(2), denotedFn(3). At this point we start to notice a pattern. Since we start with the column labled Fn(0), each column is labeled one less than the actual column it is in, and is a convolution of the first column and the previous column. Here forward, we say that the mth column, donoted Fn(m−1), is the convolution of Fn(0) and Fn(m−2) thus giving us the array in Table 1[KOS14].

Notice we have the table left justified and offset by one row. As mentioned in Chapter 3, depending on where the first generating function starts, we will have the tables off-set by the numerator of the generating function. In this case, we have the generating function[KOS14] gn(x) = x

n

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F(0) F(1) F(2) F(3) F(4) 1

1 1

2 2 1

3 5 3 1

5 10 9 4 1

8 20 22 14 5

13 38 51 40 20

21 71 111 105 65 34 130 233 256 190 55 235 474 594 511 89 420 942 1324 1295 144 744 1836 2860 3130 233 1308 3522 6020 7285

Table 4.1: Fibonacci Convolution Triangle

4.1.1 Sum of Pascal’s Triangle Rows

One of the first observations Koshy talks about in his book is that the sum of the rows of Pascal’s triangle are the Fibonacci numbers[KOS14]. In order to see this, we must first see the set up of Pascal’s triangle. Recall from Chapter 1, we have Pascal’s triangle left justified and offset by one.

Table 4.2: Pascal’s Triangle Left Justified n

0

n

1

n

2

n

3

n

4

n

5

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

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made up of the sum of the previous two numbers[KOS18]. We have the sequence 1, 1, 2, 3, 5, 8, ... as the first few terms of the sequence[KOS18]. Notice, if we were to continue the sequence, we would have the same numbers as seen in the right-most column.

Table 4.3: Pascal’s Triangle Left Justified n 0 n 1 n 2 n 3 n 4 n 5 Sum 1 1 1 1

1 1 2

1 2 3

1 3 1 5

1 4 3 8

1 5 6 1 13

1 6 10 4 21

1 7 15 10 1 34

1 8 21 20 5 55

1 9 28 35 15 1 89

At this moment, we can see if what we see is true. Recall from Chapter 4, we said that the generating function for Pascal’s triangle isg(x) = (1xnx)−n1−1. Sincegr(x) = x

2r

(1−x)r+1 is the generating function for Pascal’s triangle[KOS18] where each column starts two rows below the start of the previous column. We can say that

X

r=0

gr(x) =

X

r=0

x2r (1−x)r+1 = 11x

X

r=0 ( x

2

1−x) r.

Recall that the sum of a geometric series is as follows

X

n=0

a0(r)n = 1a0r.

Now, applying that same concept to this problem, we can easily see that

X

r=0 x2r

(1−x)r+1 = 1 1−x2

1−x

, which forces the following

X

r=0

gr(x) = 1−1x· 1 1− x2

1−x

= (1x)1x2 = 1x1x2

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4.1.2 Pascal Meets Fibonacci

One of the other proofs Koshy talks about is the convolution of each row of Pascal’s triangle make up the Fibonacci convolution triangle[KOS14]. Let us start with Pascal’s triangle left justified. Notice how each of the empty spaces above each column are filled with zeros.

Table 4.4: Pascal’s Triangle Left Justified n 0 n 1 n 2 n 3 n 4 n 5

1 0 0 0 0 0

1 1 0 0 0 0

1 2 1 0 0 0

1 3 3 1 0 0

1 4 6 4 1 0

1 5 10 10 5 1

Each of these columns are going to be the multipliers to Pascal’s triangle offset by two rows[KOS14]. Notice each of the columns start two entries below the start of the previous column. We can see the fixed Pascal’s triangle that will be used to multiply each of the entries of the columns.

Table 4.5: Pascal’s Triangle Left Justified and Offset by Two Entries n 0 n 1 n 2 n 3 n 4 n 5 1 1 1 1 1 2

1 3 1

1 4 3

1 5 6 1

1 6 10 4

1 7 15 10 1

1 8 21 20 5

1 9 28 35 15 1

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Table 4.6: Pascal’s Triangel Offset by Two Entries

1 1 1 1 1 1

1 1 1 1 1 2 1 3 1 1 4 3

1 5 6 1

1 6 10 4 1 7 15 10 1 1 8 21 20 5 1 9 28 35 15 1

As you see here the first entry in the first column will be multiplied with all the entries in the first column, the second entry in the first column will be multiplied to all the entries in the second column, the third entry in the first column will be multiplied to every entry in the third column, ..., the nthentry in the first column will be multiplied to every entry in thenthcolumn, ... , then all entries in each row will be added together. This will make up the first column of the Fibonacci convolution triangle[KOS14].

Table 4.7: First Column of Fibonacci Convolution Triangle

1 1 1 1 1 sum

1·1 1

1·1 1

1·1 + 1·1 2

1·1 + 1·2 3

1·1 + 1·3 + 1·1 5

1·1 + 1·4 + 1·3 8

1·1 + 1·5 + 1·6 + 1·1 13 1·1 + 1·6 + 1·10 + 1·4 21 1·1 + 1·7 + 1·15 + 1·10 + 1·1 34

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are going to apply the same steps as the previous table to the next table, but this time we will use the second column of Pascal’s triangle. We are going to use the first entry in the second column to multiply with all the entries in the first column, the second entry in the second column will be multiplied to all the entries in the second column, the third entry in the second column will be multiplied to every entry in the third column, ..., the nth entry in the second column will be multiplied to every entry in thenth column, ... , then all entries in each row will be added together. This will make up the second column of the Fibonacci convolution triangle[KOS14].

Table 4.8: Second Column of Fibonacci Convolution Triangle

0 1 2 3 4 sum

0·1 0

0·1 0

0·1 + 1·1 1

0·1 + 1·2 2

0·1 + 1·3 + 2·1 5

0·1 + 1·4 + 2·3 10

0·1 + 1·5 + 2·6 + 3·1 20 0·1 + 1·6 + 2·10 + 3·4 38 0·1 + 1·7 + 2·15 + 3·10 + 4·1 71

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Table 4.9: Third Column of Fibonacci Convolution Triangle

0 0 1 3 6 sum

0·1 0

0·1 0

0·1 + 0·1 0

0·1 + 0·2 0

0·1 + 0·3 + 1·1 1

0·1 + 0·4 + 1·3 3

0·1 + 0·5 + 1·6 + 3·1 9 0·1 + 0·6 + 1·10 + 3·4 22 0·1 + 0·7 + 1·15 + 3·10 + 6·1 51

The first four entries of this column are made up of zeros. This gives us more evidence that each of the columns generated in this way are offset by two rows. The other entries are the same entries of the third column of the Fibonacci convolution triangle[KOS14]. Let us try this one last time to truly see the pattern. Let us use the first entry in the fourth column to multiply with all the entries in the first column, the second entry in the fourth column will be multiplied to all the entries in the second column, the third entry in the fourth column will be multiplied to every entry in the third column, ..., the nth entry in the fourth column will be multiplied to every entry in the nth column, ... , then all entries in each row will be added together. This will make up the fourth column of the Fibonacci convolution triangle[KOS14].

Table 4.10: Fourth Column of Fibonacci Convolution Triangle

0 0 0 1 4 sum

0·1 0

0·1 0

0·1 + 0·1 0

0·1 + 0·2 0

0·1 + 0·3 + 0·1 0

0·1 + 0·4 + 0·3 0

0·1 + 0·5 + 0·6 + 1·1 1 0·1 + 0·6 + 0·10 + 1·4 4 0·1 + 0·7 + 0·15 + 1·10 + 4·1 14

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we just created by the multipliers and sums of the rows. This table is the Fibonacci convolution triangle offset by two rows[KOS14].

Table 4.11: Fibonacci Convolution Triangle F(0) F(1) F(2) F(3)

1 0 0 0

1 0 0 0

2 1 0 0

3 2 0 0

5 5 1 0

8 10 3 0

13 20 9 1

21 38 22 4

34 71 51 14

It seems as if we have constructed the Fibonacci convolution triangle[KOS14] in this way, however, we need to prove this to be true. Since gr(x) = x

r

(1−x)r+1 is the generating function for Pascal’s triangle where each column starts two rows below the start of the previous column, and we input 1x2x we get the following

gr(1x2x) = (

x2

1−x) r

(1−x2 1−x)r+1

= (

x2

1−x) r

(1−x2

1−x)r+1

·(1(1x)x)rr+1+1 = ((1(1x)−x)xx22)rr+1 = (1(1xx)xx2)2rr+1.

Lastly, by dividing both sides by (1−x), we have 1

1−x·gr( x2 1−x) =

x2r

(1−x−x2)r+1

Which is the generating function for the rth column of the Fibonacci convolution triangle[KOS14].

4.1.3 Fibonacci Meets Pell

In his book, Koshy talks about a discovery where the sum of the rows in the Fibonacci convolution triangle make up the Pell sequence[KOS14]. Recall the sequence of Pell numbers[KOS11] as the sequence of 1, 2, 5, 12, 29, 70, ... The table below gives a visual of the situation.

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Table 4.12: Sums of Fibonacci Convolution Triangle Result in Pell Numbers F(0) F(1) F(2) F(3) F(4) F(5) F(6) F(7) Row Sums

1 1

1 1 2

2 2 1 5

3 5 3 1 12

5 10 9 4 1 29

8 20 22 14 5 1 70

13 38 51 40 20 6 1 169

21 71 111 105 65 27 7 1 408

rth column of the Fibonacci convolution triangle is modeled by the generating function gr(x) = x

r

(1−x−x2)r+1 then the sum of the rows of the Fibonacci convolution triangle is modeled by[KOS14]

X

r=0

gr(x) =

X

r=0

xr (1−x−x2)r+1 = 1x1x2

X

r=0

xr (1−x−x2)r = 1x1x2

X

r=0

( x

1−x−x2) r.

Recall that the sum of a geometric series is as follows

X

n=0

a0(r)n = 1a0−r.

Which means we have the following equivalence

X

r=0

(1xxx2)r = 1 1x

1−x−x2

and thus we have the following

X

r=0

gr(x) = 1x1x2 ·1 1x

1−x−x2

= (1x1x2)x = 12x1x2

which is the generating function of the Pell numbers[KOS14].

4.1.4 Fibonacci meets Tribonacci

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Table 4.13: Sum of Rows in Fibonacci Convolution Triangle are Tribonacci Numbers F(0) F(1) F(2) Row Sums

1 1

1 1

2 2

3 1 4

5 2 7

8 5 13

13 10 1 24

21 20 3 44

Notice that the columns are offset by 3 rows. Even though the pattern looks to be this exact pattern, we must prove the conjecture. Since the rth column of the Fibonacci convolution triangle is generated by the functiongr(x) = x

r

(1−x−x2)r+1, then we can easily use the functiongr(x) = (1xx3xr2)r+1 to generate the Fibonacci convolution triangle where each column starts three rows below the the start of the previous column[KOS14]. Now, the sum of each of the rows.

X

r=0

gr(x) =

X

r=0

x3r (1−x−x2)r+1

= 1x1x2

X

r=0

x3r (1−x−x2)r = 1x1x2

X

r=0

( x

3

1−x−x2) r

= 1

1− x3

1−x−x2

= 1x1x2 · 1 1− x3

1−x−x2 = (1x1x2)x3 = 1x1x2x3

which is the generating function of the Tribonacci numbers[KOS14].

4.2

Tribonacci Convolution Triangles

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of Tn(0) and Tn(1), we say Tn(2). The fourth column results in the convolution of Tn(0) and Tn(2), we say Tn(3). Noticing the same symmetry from the Fibonacci convolution triangle, we see that each column is labeled one less than the actual column it is in, and is a convolution of the first column and its previous column. Here forward, we say that the mth column, denoted Tn(m−1), is the convolution of Tn(0)·Tn(m−2), thus forming the array in the Table below[KOS14].

Table 4.14: Tribonacci Convolution Triangle T(0) T(1) T(2) T(3) T(4)

1 1 1 1 1

1 2 3 4 5

2 5 9 14 20

4 12 25 44 70

7 26 63 125 220

13 56 153 336 646 24 118 359 864 1800 44 244 819 2144 4810 81 499 1830 5174 12430 149 1010 4018 12200 31240

4.2.1 Tribonacci Meets Fibonacci

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Table 4.15: First Column of Fibonacci Convolution Triangle

1 -1 1 Sums

1·1 1

1·1 1

1·2 2

1·4 - 1·1 3

1·7 - 1·2 5

1·13 - 1·5 8

1·24 - 1·12 + 1·1 13 1·44 - 1·26 + 1·3 21

This makes up the first column in the Fibonacci convolution triangle[KOS14]. Doing the same action using the second column of Pascal’s triangle, we have the table below.

Table 4.16: Second Column of Fibonacci Convolution Triangle

0 1 -2 Sums

0·1 0

0·1 0

0·2 0

0·4 + 1·1 1

0·7 + 1·2 2

0·13 + 1·5 5

0·24 + 1·12 - 2·1 10 0·44 + 1·26 - 2·3 20

This makes up the second column of the Fibonacci convolution triangle[KOS14]. Again, we construct the third column in the same way.

Table 4.17: Third Column of Fibonacci Convolution Triangle

0 0 1 Sums

0·1 0

0·1 0

0·2 0

0·4 + 0·1 0

0·7 + 0·2 0

0·13 + 0·5 0

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At this point we can say we have enough evidence to gather the columns and call it the Fibonacci convolution triangle[KOS14]. The table below is the columns we constructed in the previous steps and gathered in their respective order.

Table 4.18: Fibonacci Convolution Triangle F(0) F(1) F(2)

1 0 0

1 0 0

2 0 0

3 1 0

5 2 0

8 5 0

13 10 1

21 20 3

Here, we can visually see the pattern forming, but we have to rule out the fact that it was just coincidence. We start with Pascal’s generating function[KOS18]. Since gr(x) =

xr

(1−x)r+1 is the generating function for Pascal’s triangle where each column starts one row below the start of the previous column. Since we are alternating between positive and negative numbers, we must look at the negative input of x.

gr(−x) = (1((x)x))rr+1 = (1+x)(−1)rrx+1r

Recall that the generating function for the Tribonacci sequence[KOS18] in this problem is g(x) =(1xxx32x3). Let us input the Tribonacci sequence into Pascal’s triangle generating function[KOS14].

gn( x 3

(1−x−x2x3)) =

( x3 (1−x−x2−x3))

n

(1+ x3

(1−x−x2−x3))n+1

= (

x3

1−x−x2−x3)n (1+ x3

(1−x−x2−x3) n+1 ·

(1−x−x2x3)n+1 (1−x−x2x3)n+1 = (1(1xxx−2x2x−3x+x3)x3)3nn+1

= (1−x(1−x2xxx32)()−n+11)nx3n By dividing both sides by (−1)n(1xx2x3) we get

gn(x) = x 3n

(1−x−x2)n+1

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Chapter 5

Our Findings

One of the most difficult parts after finding a pattern was finding a way to prove what I had found. The book I had been looking at worked with the column generating functions[KOS14]. Then I came across a journal article from Hoggat and Bicknell that found the convolution triangles in row generating functions[HB72]. Hoggat and Bicknell proved their finding in a different way, however, I believe I have made the process more simple. Let us start with a brief explanation of what we need in order to find the terms themselves. We need a way to write the rows so that we can find the individual terms. The way Hoggat and Bicknel[HB72] found it was through the function g(x) = (11x)n.

Writing the function in this way will open the opportunity to find the mth term in that row. Later in this chapter, we will prove the relation Rn = (1N−x)nn, where Nn is the numerator of the row generating function[HB72]. This table comes from the generating function expansion since the mth term of a generating function comes from

g(x) = (11x)n

= (1 +x+x2+· · ·+xm+· · ·)n

= (1 + 1+n1−1x+ 2+n2−1x2+· · ·+ m+nm−1xm+· · ·) Therefore, the term itself is the coefficent Cn(m)= m+nm−1

Nn

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Table 5.1: Term Coefficients for Row Generating Functions C(0) C(1) C(2) · · · C(m) R1 1·N1 1+11−1

·N1 2+12−1

·N1 · · · m+1m−1

·N1 R2 1·N2 1+21−1

·N2 2+22−1

·N2 · · · m+2m−1

·N2 R3 1·N3 1+31−1

·N3 2+32−1

·N3 · · · m+3m−1

·N3 ..

. ... ... ... . .. ...

Rn 1·Nn 1+n1−1

·Nn 2+n2−1

·Nn · · · m+nm−1

·Nn

5.1

Fibonacci Convolution Triangle

The first thing we need to show is that there exists row generating functions rather than column generating functions. This is easily proved through st induction[POL81].

5.1.1 Row Generators

We can start the process by seeing the difference and similarities in rows. We can start with the very first row. Notice it is a row of ones. Since we have proven this relation in a previous chapter[KOS18], we can say R1 = 1−1x. The second row is a row of the natural numbers, which is similar to the second row of Pascal’s triangle[KOS18], which would make R2 = (11x)2. We start to notice a pattern in the third row[HB72]. If we were to offset the third row by one column, and add it to the previous two, we would get the third row as shown below[HB72].

R1 1 1 1 1 1

R2 1 2 3 4 5

+ R3 2 5 9 14

R3 2 5 9 14 20

This pattern worked for the third row, let us see if it works out for the fourth row[HB72].

R2 1 2 3 4 5

R3 2 5 9 14 20

+ R4 3 10 22 40

R4 3 10 22 40 65

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R3 2 5 9 14 20

R4 3 10 22 40 65

+ R5 5 20 51 105

R5 5 20 51 105 190

.

At this point, we can say that each row, starting with the third row comes from the pattern Rn= x·Rn+Rn−1+Rn−2 which would give us Rn = Rn−11+R−xn−2. Let us find the first few rows[HB72].

R1 = 1−1x R2 = (11x)2

R3 = R21−+R1x =

1 (1−x)2+

1 1−x

1−x = (11+(1x)(1x)x)2 = (12x)x3 R4 = R31+R2x

= 2−x

(1−x)3+ 1 (1−x)2 1−x = (2(1x)+(1x)(1x)x)3 = (13−2xx)4 R5 = R41+R3x

=

3−2x

(1−x)4+ 2−x

(1−x)3 1−x

= (3−2x)+(2(1x)(1x)(1x)4−x) = 3−2x+2(1x)3x+x5 2 = 5−(15x+xx)52

Notice that the degree of the denominator of thenth row isn. If we were to separate the numerator from the denominator, we would haveRn= (1Nnx)n for some numeratorNn. If

we were to take the numerator as the sum of the two numerators of the rows, we notice that we have the sum of two quotients with different degrees in their denominator[HB72]. This seems to follow the relation of Rn= (1N−nx)n =

Nn−1+(1−x)Nn−2

(1−x)n where the numerator

of the nth row isNn =Nn−1+ (1−x)Nn−2. We will use this relation to help prove our findings.

5.1.2 Sum of Rows for the mth Term

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.

Table 5.2: Sum of Rows F(0) F(1) F(2) F(3) F(4)

R1 1 1 1 1 1

R2 1 2 3 4 5

R3 2 5 9 14 20

R4 3 10 22 40 65

R5 5 20 51 105 190

R6 8 38 111 256 511

For each entry in the mth column, we have the sum of the two rows above it, up the the mth entry of their respective columns resulting in the mth term that we speak of. For example, notice in table 5.2, we have the sum of the two rows above the entry 40 resulting in 40. We can prove this with induction. Recall table 5.1 which shows the mth term in the nth column through the coefficients of the generating functions. Using those coefficients, we can solve this relation by induction. In order to prove this by induction, we must prove the two different shifts. A horizontal shift in terms and a vertical shift in terms. Before we start, let’s note the following equivalence:

Cn(m) = m+nm−1

Nn

= m+nm−1·[Nn−1+ (1−x)Nn−2]

= m+nm−1Nn−1+ m−1+nm −1

Nn−2 = m+nm−1Nn−1+ m+nm−2

Nn−2

We know by the table that C3(4)= 20. The following is an example of the relation above: C3(4) = 4+34−1

N3 = 4+34−1

·[N3−1+ (1−x)N3−2] = 4+34−1

N2+ 4−1+34 −1

N1 = 64

N2+ 54

N1 = 15N2+ 5N1 = 15+5 = 20

We want to prove the mth term in thenth row is the sum of the two rows above it. We will first prove the horizontal shift in terms.

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C3(0) = 0+30−1·N3 = 20·N3 = 1·N3

Now that we have the LHS, let us see what the result of the RHS is. 0+1−1

0

·N1+ 0+20−1

·N2 = 0+20−1

·N3 = 1·N3

Since we got the same results for the two sides, we have proven the first term to be true. Now, we have to check the entire rule.

Step 2: Assume Cn(k) is true, prove Cn(k+1) We take the following to be true.

(1 + 1+n−12−1+ 2+n−22−1+· · ·+ k+n−k2−1)·Nn−2

+ = k+nk−1·Nn (1 + 1+n−11−1

+ 2+n−21−1

+· · ·+ k+n−k1−1

)·Nn−1 If we simplify, we get

(1 + n−12+ n−21+· · ·+ k+nk−3)·Nn−2

+ = k+nk−1

·Nn (1 + n−11+ n2+· · ·+ k+nk−2)·Nn−1

We must now prove for the termCn(k+1). The RHS is fairly easy to prove.

Cn(k+1) = k+1+nk+1−1·Nn = k+nk+1·Nn

Now we prove that the sum of the two rows above Cn(k+1) results inCn(k+1).

(1 + 1+n−12−1

+ 2+n−22−1

+· · ·+ k+n−k2−1

+ k+1+nk+1−2−1

)·Nn−2+ (1 + 1+n−11−1+ 2+n−21−1+· · ·+ k+n−k1−1+ k+1+nk+1−1−1)·Nn−1

When we simplify we get

(1 + n−12+ n−21+· · ·+ k+nk−3+ k+nk+1−2)·Nn−2+

(1 + n−11 + n2

+· · ·+ k+nk−2

+ k+nk+1−1

)·Nn−1

By the rule of induction, we substitute and get

k+n k

·Nn+ k+nk+1−1

·Nn−1+ k+nk+1−2

·Nn−2

(47)

k+n−1 k

·Nn+ k+nk+1−1

·Nn

Lastly, by Pascal’s identity, we have k+n

k+1

·Nn Since k+nk+1

·Nn = k+nk+1

·Nn we can say that the sum of the two rows above the Cn(m) term results in Cn(m) for a horizontal shift in terms.

Now that we proved this relation to be true in the rows, let’s see the proof by induction for the shifts in columns. In this case, we will prove the first column to be true. Note that the first column is the Fibonacci sequence. Knowing that every entry is the sum of the two entries above it in the first column, we can shift horizontally to prove this conjecture works for every entry in the Fibonacci convolution triangle.

Step 1: Prove the first term of the series. In this case the first term we can apply the rule to is C3(0). We know this to be true since we have proved this for the horizontal shifts in the array. Let’s move on to step 2.

Step 2: Assume Ck(0) is true, proveCk+1(0)

We take the following to be true. 0+k−1

0

Nn = 0+k−01−1

Nk−1+ 0+k−02−1

Nk−2

When we simplify we get

k−1 0

Nk = 0+k0−2

Nk−1+ 0+k0−3

Nk−2

Now we proveCk+1(0) to be the sum of the two numbers above it. Let’s start with the LHS.

Ck+1(0) = 0+k+10 −1 Nk+1

When we simplify we get

k 0

Nk+1

(48)

k 0

·[Nk+ (1−x)Nk−1] = k0Nk+ k−01

Nk−1

When we simplify we get

1·Nk+ 1·Nk−1 = Nk+Nk−1

Now that we have the result for the LHS, lets see the result of the RHS.

Ck+1(0) = 0+k0−1

Nk+ 0+k−01−1

Nk−1 = k−01

Nk+ k−02

Nk−1 = 1·Nk+ 1·Nk−1 = Nk+Nk−1

Since we proved the horizontal shift to be true, and since the LHS=RHS, we know that we can use this conjecture to generate the entire Fibonacci convolution triangle. Therefore, the sum of the two rows up to the mth entry aboveC(m)

n is Cn(m).

5.1.3 The Sum of Distinct Term for the mth Term

If you were to offset the columns by two rows, we would have the following table.

Table 5.3: Sum of Terms Offset by Two Rows F(0) F(1) F(2) F(3)

1 1

2 1

3 2

5 5 1

8 10 3

13 20 9 1

21 38 22 4

34 71 51 14

55 130 111 40

89 235 233 105

(49)

below to show us where those numbers came from.

Table 5.4: Sum of Terms F(0) F(1) F(2) F(3) F(4)

1 1 1 1 1

1 2 3 4 5

2 5 9 14 20

3 10 22 40 65

5 20 51 105 190 8 38 111 256 511

Here we can see that it generates the first few entries, but does it generate the entire array? We can prove this through induction in two cases. First for the horizontal shift, then for the vertical shift.

We want to prove thatCn(m) =Cn(m−1)+Cn(m)−1+C (m)

n−2. Let us first see the case of the horizontal shift.

Step 1: Prove the first term which, in this case isC3(1) We can start by seeing the result of the LHS.

C3(1) = 1+31−1·N3 = 31·N3

Now, let us see the result for the RHS. C3(0) +C2(1) +C1(2) = 0+30−1

·N3+ 1+21−1

·N2+ 2+12−1

·N1 = 20·N3+ 1+21−1

·N3 = 20·N3+ 21

·N3 = 31·N3

Since 31

·N3= 31

·N3, we can say that the relation works for the first term. Now that we have finished the first step, we can now move on to step two.

Step 2: Assume Cn(k) is true, prove Cn(k+1) We take the following to be true.

Cn(k) = Cn(k−1)+Cn(k)−1+C (k) n−2 k+n−1

k

·Nn = k−1+nk1−1

·Nn+ k+n−k1−1

·Nn−1+ k+n−k2−1

·Nn−2 k+n−1

k

·Nn = k+nk−−12

·Nn+ k+nk−2

·Nn−1+ k+nk−3

·Nn−2

(50)

Cn(k+1) = k+1+nk+1−1

·Nn = k+nk+1

·Nn

Now that we have the LHS, the next thing we do is see if we get the same result for the RHS.

Cn(k−1)+Cn(k)−1+C (k) n−2 = k+1+nk −2·Nn+ k+1+nk+1−2

·Nn−1+ k+1+nk+1−3

·Nn−2

When we simplify we get

= k+nk−1

·Nn+ k+nk+1−1

·Nn−1+ k+nk+1−2

·Nn−2

By substituting the nth numerator we get

= k+nk−1·Nn+ k+nk+1−1

·Nn

When we simplify we get

= k+nk+1·Nn

Since k+nk+1

·Nn= k+nk+1

·Nn, we can say the relation holds true for the horizontal shift. If we were to do the proof for the vertical shift, we can see that if we prove this relation for the first column, we can prove it for the entire array. Notice, if we were to take the two terms above the first term in the nth row, we have the exact proof we proved in the previous section. Note, that the entry to the left of any term in the first column is zero. Therefore, we have already proven the relation to be true for the vertical shift. Therefore, each term in the array can be made up of the two numbers above it and the one to the left of it.

5.2

Tribonacci Convolution Triangle

(51)

made it more difficult to understand[HB72]. I used the row generating function they found and used induction to prove both their findings, and my own. We can start with the row generating function[HB72].

5.2.1 Row Generators

Picking up parts from the Fibonacci row generating function[HB72], we can see that the first three rows of the Tribonacci convolution triangle are the same as the ones we found in the Fibonacci convolution triangle. We start seeing a difference in the 4th row. Similar to the Fibonacci convolution triangle, if we were to offset the fourth row by one column, and add it to the previous three, we would get the third row[HB72]. Below is a visual of what I am stating.

R1 1 1 1 1 1

R2 1 2 3 4 5

R3 2 5 9 14 20

+ R4 4 12 25 44

R4 4 12 25 44 70

This pattern gives us the fourth row. Let’s see if this works for the fifth row.

R2 1 2 3 4 5

R3 2 5 9 14 20

R4 4 12 25 44 70

+ R5 7 26 63 125

R5 7 26 63 125 220

This gives us the fifth row of the Tribonacci convolution triangle[HB72]. Let’s give it one more try before we see the pattern.

R3 2 5 9 14 20

R4 4 12 25 44 70

R5 7 26 63 125 220

+ R6 13 56 153 336

R6 13 56 153 336 646

At this point, it is safe to see the pattern[HB72] we have found to being Rn =x·Rn+ Rn−1+Rn−2+Rn−3which would give usRn= Rn−1+R(1nx)2+Rn−3. Similar to the Fibonacci convolution triangle we can say thatRn= (1N−x)nn for some numeratorNn. This will allow us to use the coefficient chart in the beginning of the chapter. It also seems to follow the relation Rn = (1Nx)nn =

Nn−1+(1−x)Nn−2+(1−x)2Nn−3

(1−x)n where the numerator of the nth row

(52)

5.2.2 Sum of the Rows for the mth Term

Each of the terms in the Tribonacci convolution triangle are the result of the sum of the three rows above it up to the that term[HB72]. The table below is a good representation of the situation.

Table 5.5: Sum of Rows T(0) T(1) T(2) T(3) T(4)

1 1 1 1 1

1 2 3 4 5

2 5 9 14 20

4 12 25 44 70

7 26 63 125 220 13 56 153 336 646

Notice that 125 = (1+2+3+4)+(2+5+9+14)+(4+12+25+44), which is the sum of the three rows above 125. This generates the entire Tribonacci convolution triangle. This can be easily proven by induction as shown below[HB72]. Before we start, let’s note the following equivialence:

Cn(m) = m+nm−1

Nn

= m+nm−1·[Nn−1+ (1−x)Nn−2+ (1−x)2Nn−3] = m+nm−1

Nn−1+ m−1+nm −1

Nn−2+ m−2+nm −1

Nn−3 = m+nm−1

Nn−1+ m+nm−2

Nn−2+ m+nm−3

Nn−3

We want to prove the mth term in the nth row is the sum of the three rows above it. Step 1: Prove the first term of the series. In this case, the first term we can apply will be C4(0). Let’s start with finding the LHS.

C4(0) = 0+40−1·N4 = 30

·N4 = 1·N4

Now that we have the result for the LHS, let’s see what we get for the RHS. C1(0)+C2(0)+C3(0) = 0+10−1N1+ 0+20−1

N2+ 0+30−1

N3 = 0+10−1N4

= 1·N4

Since N4 =N4, we can say that the conjecture is true for the first term. The next thing we have to do is prove the general case.

(53)

(1 + 1+n−13−1+ 2+n−23−1+· · ·+ k+n−k3−1)·Nn−3+

k+n−1 k

Nn = (1 + 1+n−12−1

+ 2+n−22−1+· · ·+ k+n−k2−1)·Nn−2+

(1 + 1+n−11−1

+ 2+n−21−1

+· · ·+ k+n−k1−1

)·Nn−1 By simplifying, we get

(1 + n−13+ n−22+· · ·+ k+nk−4)·Nn−3+

k+n−1 k

Nn = (1 + n−12

+ n−21

+· · ·+ k+nk−3

)·Nn−2+ (1 + n−11+ n2+· · ·+ k+nk−2)·Nn−1

The next thing we do is prove the following term. Let us start with the LHS. Cn(k+1) = k+1+nk+1−1

·Nn = k+nk+1·Nn

Now that we have the LHS, let’s see what the result of the RHS is.

(1 + 1+n−13−1+ 2+n−23−1+· · ·+ k+n−k3−1+ k+1+nk+1−3−1)·Nn−3+ (1 + 1+n−12−1

+ 2+n−22−1

+· · ·+ k+n−k2−1

+ k+1+nk+1−2−1

)·Nn−2+ (1 + 1+n−11−1+ 2+n−21−1+· · ·+ k+n−k1−1+ k+1+nk+1−1−1)·Nn−1 After simplifying, we get

(1 + n−13+ n−22+· · ·+ k+nk−4+ k+nk+1−3)·Nn−3+ (1 + n−12+ n−21+· · ·+ k+nk−3+ k+nk+1−2)·Nn−2+

(1 + n−11+ n2+· · ·+ k+nk−2+ k+nk+1−1)·Nn−1 By the induction hypothesis, we substitute and get

k+n−1 k

Nn+ k+nk+1−1

Nn−1+ k+nk+1−2

Nn−2+ k+nk+1−3

Nn−3

By the definition of thenth numerator, we substitute and get

k+n−1 k

Nn+ k+nk+1−1

Nn

Lastly, by Pascal’s identity, we have

k+n k+1

Nn

Since k+nk+1

Nn= k+nk+1

(54)

Now let’s prove the relation for a vertical shift.

Step 1: Prove the first term of the series. In this case the first term we can apply the rule to is C4(0). We know this to be true since we have proved this for the horizontal shifts in array. Let’s move on to step 2.

Step 2: Assume Ck(0) is true, proveCk+1(0) We take the following to be true.

0+k−1 0

Nk = 0+k−01−1

Nk−1+ 0+k−02−1

Nk−2+ 0+k−03−1

Nk−3 k−1

0

Nk = k−02

Nk−1+ k−03

Nk−2+ k−04

Nk−3

Now we prove Ck+1(0) to be the sum of the two numbers above it. Let’s start with the LHS. Ck+1(0) = 0+k+10 −1

Nk+1

By simplifying, we get

k 0

Nk+1

By substituting the nth numerator, we get

k 0

·[Nk+ (1−x)Nk−1+ (1−x)2Nk−2]

By distributing, we get

k 0

Nk+ k−01

Nk−1+ k−02

Nk−2

By simplifying, we get

1·Nk+ 1·Nk−1+ 1·Nk−2 = Nk+Nk−1+Nk−2

Now that we have the result for the LHS, lets see the result of the RHS. Ck+1(0) = 0+k0−1

Nk+ 0+k−01−1

Nk−1+ 0+k−02−1

Nk−2

(55)

k−1 0

Nk+ k−02

Nk−1+ k−03

Nk−2 = 1·Nk+ 1·Nk−1+ 1·Nk−2

= Nk+Nk−1+Nk−2

Since the LHS=RHS, and since we proved the horizontal shift, we know the sum of the three rows up to the mth entry above Cn(m) isCn(m).

5.2.3 The Sum Distinct Terms for the mth Term

In order to see the actual pattern, we must first see the Tribonacci convolution triangle as left justified, and offset by three rows as in the table below. This allows us to see the distinct terms that make up the mth term in the nth row.

Table 5.6: Sum of Terms Offset by Three Rows T(0) T(1) T(2) T(3) T(4)

1 1 2

4 1

7 2

13 5

24 12 1

44 26 3

81 56 9

149 118 25 1 274 244 63 4

504 499 153 14

927 1010 359 44 1 1705 2027 819 125 5

In this case, we see 125 = 44 + 14 + 4 + 63. It is easy to see the pattern forming, but we should see these numbers on the simply left justified Tribonacci convolution triangle as show below.

At this point, we want to prove Cn(m)=C(m

−1)

n +Cn(m)−1+C (m) n−2+C

(m)

n−3. Let’s prove this by form of induction. We want to prove the mth term in the nth row is the sum of the three entries above it, and the one to the left.

Figure

Figure 1.7: Pascal’s Triangle (constructing fifth row)

Figure 1.7:

Pascal’s Triangle (constructing fifth row) p.14
Figure 1.9: Pascal’s Triangle

Figure 1.9:

Pascal’s Triangle p.15
Table 1.1: Pascal’s Triangle Left Justified��������������

Table 1.1:

Pascal’s Triangle Left Justified�������������� p.16
Table 1.2: Fibonacci Convolution Triangle

Table 1.2:

Fibonacci Convolution Triangle p.16
Table 1.3: Tribonacci Convolution Triangle

Table 1.3:

Tribonacci Convolution Triangle p.17
Table 4.2: Pascal’s Triangle Left Justified������������

Table 4.2:

Pascal’s Triangle Left Justified������������ p.30
Table 4.1: Fibonacci Convolution Triangle

Table 4.1:

Fibonacci Convolution Triangle p.30
Table 4.3: Pascal’s Triangle Left Justified������������

Table 4.3:

Pascal’s Triangle Left Justified������������ p.31
Table 4.4: Pascal’s Triangle Left Justified������������

Table 4.4:

Pascal’s Triangle Left Justified������������ p.32
Table 4.5: Pascal’s Triangle Left Justified and Offset by Two Entries������������

Table 4.5:

Pascal’s Triangle Left Justified and Offset by Two Entries������������ p.32
Table 4.6: Pascal’s Triangel Offset by Two Entries

Table 4.6:

Pascal’s Triangel Offset by Two Entries p.33
Table 4.7: First Column of Fibonacci Convolution Triangle

Table 4.7:

First Column of Fibonacci Convolution Triangle p.33
Table 4.11: Fibonacci Convolution Triangle

Table 4.11:

Fibonacci Convolution Triangle p.36
Table 4.12: Sums of Fibonacci Convolution Triangle Result in Pell Numbers

Table 4.12:

Sums of Fibonacci Convolution Triangle Result in Pell Numbers p.37
Table 4.13: Sum of Rows in Fibonacci Convolution Triangle are Tribonacci Numbers

Table 4.13:

Sum of Rows in Fibonacci Convolution Triangle are Tribonacci Numbers p.38
Table 4.14: Tribonacci Convolution Triangle

Table 4.14:

Tribonacci Convolution Triangle p.39
Table 4.17: Third Column of Fibonacci Convolution Triangle

Table 4.17:

Third Column of Fibonacci Convolution Triangle p.40
Table 4.15: First Column of Fibonacci Convolution Triangle

Table 4.15:

First Column of Fibonacci Convolution Triangle p.40
Table 4.18: Fibonacci Convolution Triangle

Table 4.18:

Fibonacci Convolution Triangle p.41
Table 5.1: Term Coefficients for Row Generating Functions

Table 5.1:

Term Coefficients for Row Generating Functions p.43
Table 5.2: Sum of Rows

Table 5.2:

Sum of Rows p.45
Table 5.3: Sum of Terms Offset by Two Rows

Table 5.3:

Sum of Terms Offset by Two Rows p.48
Table 5.4: Sum of Terms

Table 5.4:

Sum of Terms p.49
Table 5.5: Sum of Rows

Table 5.5:

Sum of Rows p.52
Table 5.6: Sum of Terms Offset by Three Rows

Table 5.6:

Sum of Terms Offset by Three Rows p.55
Table 5.7: Sum of Distinct Terms

Table 5.7:

Sum of Distinct Terms p.56
Table 6.1: Sum of Two Rows

Table 6.1:

Sum of Two Rows p.58
Table 6.2: Sum of Terms

Table 6.2:

Sum of Terms p.59
Table 6.3: Sum of Three Rows

Table 6.3:

Sum of Three Rows p.59
Table 6.4: Sum of Terms

Table 6.4:

Sum of Terms p.60
Related subjects : convolution of